Quiz No. 2 With Solution Properties of Soil

CE 412  –  Geotechnical  Geotechnical Engineering I (Soil Mechanics) Quiz No. 2 Properties of Soil

Name :

Answer Key

Student No. :

Course &Sec :

Rating :

Solve the following problems and kindly show your solution. 1. An undisturbed sample of fine sand is tested in the laboratory and found to have a dry mass of 3.63 kg, a total volume of 0.00198 m 3 and a specific gravity of 2.70. Other laboratory tests are performed to determine the maximum and minimum density of the sand. At the maximum density, it is determined that the void ratio is 0.35, at the minimum density, the void ratio is 0.95. Compute the: a. void ratio of the sand

W mg  γ = V = V    3.63 kg 9.81 s    γ = 0.00198 m  γ =17.985 kN⁄m  γ = 1 +G e γw 2.70 kN 17.985 ⁄m = 1 + e 9.81 kN⁄m =. 52 | P a g e

C E

4 1 2 - P R O P E R T IE S

O F

S O I L

b. relative density of the undisturbed sample

ea − e ×100% Dr = ea − en 0.95−0.473 Dr = 0.95−0.35 ×100%  =.% c.

dry density

m ρ = V 3.63 kg ρ = 0.00198 m  =,. ⁄ 2. A soil excavated from a borrow pit has an in-place dry unit weight of 18.59 kN/m3. In the laboratory, the values of dry maximum and minimum unit weights are 20.60 kN/m3 and 14.13 kN/m3 respectively. Compute the: a. relative density of the soil sample

1⁄ γ − 1⁄ γ Dr = 1⁄ − 1⁄ ×100%  γ  γ 1⁄14.13 − 1⁄18.59 Dr = 1⁄ − 1⁄ ×100% 14.13 20.60  =.%

53 | P a g e

C E

4 1 2 - P R O P E R T IE S

O F

S O I L

ea =0.92 and en =0.28 e − e a Dr = ea − en ×100% 0.92−e 76.39%= 0.92−0.28 ×100% =.

b. void ratio in the natural condition if

c.

specific gravity of the soil

G  γ = 1 + e γw G kN 18.59 ⁄m = 1+0.431 9.81 kN⁄m =. 3. A soil at a constant moisture content shows the following when compacted 

    , ⁄

   ,% 40

14.50

70

17.89

Determine the: a. specific gravity of soil

GMC=Se GMC e= S

54 | P a g e

C E

4 1 2 - P R O P E R T IE S

O F

S O I L

G  γ = 1 + e γw G  γ = 1 +  γw  GMC  γwG = γ 1 + S  GMC  γwG=14.501+ 0.40  ⟶ Eq.1  γwG=17.891+ GMC  ⟶ Eq.2 0.70 Eq.1=Eq.2 GMC GMC 14.501+ 0.40 =17.891+ 0.70  14.50+36.25GMC=17.89+25.56GMC 10.69GMC=3.39 GMC=0.317  γ = 1 +G γw 

G kN 14.50 ⁄m = 1 + . 9.81 kN⁄m . =.

55 | P a g e

C E

4 1 2 - P R O P E R T IE S

O F

S O I L

b. moisture content of soil

GMC=0.317 2.65MC=0.317 MC=0.1197 =.% c.

17.89 kN⁄m soil  γ = 1 +G γw  2.65 kN 14.50 ⁄m = 1 + e 9.81 kN⁄m =.

void ratio of the

4. A saturated soil used to determine the shrinkage limit has an initial volume , final volume  , mass of wet soil   and mass of dry soil  . Determine the: a. shrinkage limit

V =20.2cm

V =14.3cm m = 24 g

m = 34 g

m − m V − V    SL = m − m  ρw  − 14.3 cm g 34 g − 24 g 20.2 cm SL = 24 g − (1 ⁄cc) 24 g =.% b. shrinkage ratio

1 m SR = ρw V 24 g SR = (1 g⁄ )14.3 cm cc =. 56 | P a g e

C E

4 1 2 - P R O P E R T IE S

O F

S O I L

c.

specific gravity of soil

G =  −1 

 

G =  −1 . .  =. 5. From the table shows the summary of a liquid limit and plastic limit test.

Compute the: a. liquid limit LIQUID LIMIT TEST

Soil Sample

Wt. of wet soil (kg)

Wt. of dry soil (kg)

Water Content (%)

 No. of Blows (N)

A

9.50

6.70

41.79

28

B

7.95

5.54

43.50

20

C

8.21

5.69

44.29

18

D

12.85

8.83

45.53

13

Using Calculator Technique:

[Mode][3][4] x y 1  28 41.79  2 20 43.50 [A] Liquid Limit,LL = 25 ⏞y =.% 57 | P a g e

Note: ⏞y = [Shift][1][5][5]

C E

4 1 2 - P R O P E R T IE S

O F

S O I L

b.  plastic limit PLASTIC LIMIT TEST

Soil Sample

Wt. of wet soil (kg)

Wt. of dry soil (kg)

Water content (%)

1

9.05

7.35

29.13

2

8.66

7.01

23.54

ω + ω  PL=MCav = 2  29.13%+23.54% PL = 2 =.% c.

natural water content NATURAL WATER CONTENT

Soil Sample

Wt. of wet soil (kg)

Wt. of dry soil (kg)

Water content (%)

1

9.69

7.00

38.43

2

9.47

6.86

38.05

MCav = ω +2 ω 38.43%+38.05% MCav = 2  =.% 58 | P a g e

C E

4 1 2 - P R O P E R T IE S

O F

S O I L

d.  plasticity index

PI=LL−PL PI=42.37%−23.33% =. e.

liquidity index

MC−PL LI = LL−PL 38.24%−23.33% LI = 42.37%−23.33% =. 6. Given the tabulated data for an Atterberg Limit Test.

Compute the: a. liquid limit LIQUID LIMIT

Test  Number

1

2

3

4

39

30

23

12

Wt. of wet soil + container

22.4 g

21.19 g

21.27 g

26.12 g

Wt. of dry soil + container

20.10 g

19.28 g

19.25 g

22.60 g

Wt. of container

12.74 g

13.24 g

13.06 g

13.27 g

Water content

31.25%

31.62%

32.63%

37.73%

 No. of blows

59 | P a g e

C E

4 1 2 - P R O P E R T IE S

O F

S O I L

Using Calculator Technique:

[Mode][3][4] x y 1  23 32.63  2 30 31.62 [A] Liquid Limit,LL = 25 ⏞y =.%

Note: ⏞y = [Shift][1][5][5]

b.  plastic limit PLASTIC LIMIT

Test Number

1

2

Wt. of wet soil + container

22.12 g

21.84 g

Wt. of dry soil + container

20.42 g

20.19 g

Wt. of container

13.07 g

13.18 g

Water content

23.13%

23.54%

ω + ω  PL=MCav = 2  23.13%+23.54% PL = 2 =.%

60 | P a g e

C E

4 1 2 - P R O P E R T IE S

O F

S O I L

c.

natural water content NATURAL WATER CONTENT

Test Number

1

2

Wt. of wet soil + container

17.53 g

16.97 g

Wt. of dry soil + container

14.84 g

14.36 g

Wt. of container

7.84 g

7.50 g

38.43%

38.05%

Water content

ω + ω  MCav = 2  38.43%+38.05% MCav = 2  =.% d.  plasticity index

PI=LL−PL PI=32.32%−23.33% =. e.

liquidity index

MC−PL LI = LL−PL 38.24%−23.33% LI = 32.32%−23.33% =. 61 | P a g e

C E

4 1 2 - P R O P E R T IE S

O F

S O I L